Question

An electric current is passed through an aqueous solution of lead(II) chloride. What are the half-reactions that take place at the cathode and anode?

• A. $\text{Anode}-P{b}^{2+}+2e^{-}\to Pb$ $\text{Cathode}-C{l}^{-}\to Cl+e^{-}$
• B. $\text{Anode}-P{b}^{2+}+2e^{-}\to Pb$ $\text{Cathode}-Cl\to C{l}^{+}+e^{-}$
• C. $\text{Cathode}-P{b}^{2+}\to P{b}^{+}2e^{-}$ $\text{Anode}-C{l}^{-}\to Cl+e^{-}$
• D. $\text{Anode}-Pb\to P{b}^{2+}2e^{-}$ $\text{Cathode}-C{l}^{-}\to Cl+e^{-}$

Answer 1

The correct option is C

$\text{Cathode}-P{b}^{2+}\to P{b}^{+}2e^{-}$

$\text{Anode}-C{l}^{-}\to Cl+e^{-}$

We know oxidation happens at anode and reduction at cathode
In $PbC{l}_2$ ions are $P{b}^{2+}$ and $C{l}^{-}$ , so $P{b}^{2+}$ reduces and $C{l}^{-}$ oxidizes.
So at anode reaction is
$C{l}^{-}\to Cl+e^{-}$ (oxidation)
$P{b}^{2+}+2e^{-}\to Pb$ (reduction at cathode)